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3 years ago

How can you solve quadratic equations in one variable?

Mathematics

1 answer:

ivolga24 [154]3 years ago

6 0

you can use formular method or factorisation but formular method is easy and convinient

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A rectangular garden is 8 feet long. If you walk diagonally across the garden, you would walk 10 feet. How many feet wide is the

galina1969 [7]

By using Pythagoras theorem

width = $\sqrt{10^2 - 8^2}$ = 6 feet

3 0

3 years ago

Read 2 more answers

Derek says that the quotient of -2/7 divided by -2/21 is -1/3. Part A: What is the correct quotient? Part B: What mistake did De

Thepotemich [5.8K]

Answer:

The correct quotient is 3.

I think Derek has mistaken to divide $-\frac{2}{21}$ by $\frac{2}{7}$ .

Step-by-step explanation:

We have to check the quotient of $(-\frac{2}{7}) \textrm{ divided by }(-\frac{2}{21})$ .

Part A:

Now, $(-\frac{2}{7}) \textrm{ divided by }(-\frac{2}{21})$ .

= $\frac{-\frac{2}{7} }{-\frac{2}{21} } =3$

So, the correct quotient is 3.

Part B:

Derek says that the quotient is $-\frac{1}{3}$ .

So, I think Derek has mistaken to divide $-\frac{2}{21}$ by $\frac{2}{7}$ . (Answer)

8 0

3 years ago

Please help me in this:

Svetlanka [38]

What exactly do you need help with?!

4 0

3 years ago

I get

dybincka [34]

Answer:

Absolute minimum = 1.414

Absolute maximum = 2.828

Step-by-step explanation:

$g(x,y)=\sqrt {x^2+y^2} \ constraints: 1\leq x\leq 2 ,\ 1\leq y\leq2$

For absolute minimum we take the minimum values of $x$ and $y$ .

$x_{minimum} =1\\y_{minimum}=1\\$

Plugging in the minimum values in the function.

$g(1,1)=\sqrt {1^2+1^2}\\g(1,1) = \sqrt{1+1}\\g(1,1)=\sqrt {2}\\g(1,1)=\pm 1.414\\$

Absolute minimum value will be always positive.

∴ Absolute minimum = 1.414

For absolute maximum we take the maximum values of $x$ and $y$ .

$x_{maximum} =2\\y_{maximum}=2\\$

Plugging in the maximum values in the function.

$g(2,2)=\sqrt {2^2+2^2}\\g(2,2) = \sqrt{4+4}\\g(2,2)=\sqrt {8}\\g(2,2)=\pm 2.828\\$

Absolute maximum value will be always positive.

∴ Absolute maximum = 2.828

3 0

3 years ago

The two legs (labeled x) of the right triangle below have equal length. If the hypotenuse has length 5√2 , solve for x.

leonid [27]

Answer:

x=5

Step-by-step explanation:

Other than using the plain special aspect of a 45-45-90 triangle where the legs are x, x, and x√2, you can solve for this.

Since the two legs have equal length, they are both x. Using the pythagorean theorem:

(x^2)+(x^2)=50 (Because 5 squared is 25 and √2 squared is 2, multiplying them gives you 50).

You can add (x^2) and (x^2) because they are the same terms (x squared).

Simplifying like so gives you:

2x^2=50

Dividing by two on both sides:

x^2=25

Taking the square root of both sides:

x=5

6 0

3 years ago